The wavelengths of two sound notes in air are | vedclass

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(A) Let $v$ be the velocity of sound in air and $f_0$ be the frequency of the third note.The frequencies of the two sound notes are $f_1 = \frac{v}{\l

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(A) Let $v$ be the velocity of sound in air and $f_0$ be the frequency of the third note.The frequencies of the two sound notes are $f_1 = \frac{v}{\lambda_1} = \frac{v}{40/195} = \frac{195v}{40}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{40/193} = \frac{193v}{40}$.Since each note produces $9$ beats per second with the third note,we have:$|f_1 - f_0| = 9$ and $|f_2 - f_0| = 9$.This implies $f_1 - f_0 = 9$ and $f_0 - f_2 = 9$ (since $f_1 > f_2$).Adding these two equations:$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$$f_1 - f_2 = 18$Substituting the expressions for $f_1$ and $f_2$:$\frac{195v}{40} - \frac{193v}{40} = 18$$\frac{2v}{40} = 18$$\frac{v}{20} = 18$$v = 360 ~m/s$.

The wavelengths of two sound notes in air are $\frac{40}{195} ~m$ and $\frac{40}{193} ~m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

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